Two matrices are simultaneously diagonalizable iff they commute.
Can anything be said about the simultaneous diagonalizability if two matrices commute but with a dagger? This happens when both A and B are Hermitian.

1 Answer
1

If $A$ and $B$ are simultaneously diagonalizable, they must commute. So if in addition $AB = (BA)^\dagger$, we have $(BA)^\dagger = BA$, i.e. their product is Hermitian. Now $A$ and $B$ commute with $BA$. Thus the eigenspaces of $BA$ are invariant under both $A$ and $B$. There are two cases:

On an eigenspace $V = V_\lambda$ of $BA$ for a nonzero $\lambda$, we have $AB = \lambda I$, so
$B|_V = \lambda (A|_V)^{-1}$.